AP Chemistry Galvanic Cells

Redox reactions often take place under acidic or basic conditions in solution as well as neutral ones. Acidic redox reactionsinvolve H⁺ while basic ones involve OH⁻. To balance a redox reaction under basic conditions, balance the half reactions using H⁺, then add OH⁻ to each side to cancel any remaining H⁺ ions. If the half reactions are kept in separate beakers but connected with a wire, the electrons will travel through the path provided by the wire where they can do work. A salt bridge or porous disk can be used to balance the charges produced in the beakers by providing counter-ions. The compartment with the oxidation half-reaction is called the anode and the reduction compartment is the cathode.

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Hi, I have a question about salt bridges, if the salt is supposed to balance the charges of the solutions and therefore extend the length of time current is flowing, what happens when there are no more electrons to flow from the side that's giving them? Is that when the reaction and the current stop?

1 answer

Last reply by: Professor HovasapianTue Jun 28, 2016 9:00 PM

Post by Neil Kottaon June 28, 2016

Oh okay I think I see what you mean...could we just look at the reduction potentials in this case or no? Thank you so much for helping me

3 answers

Last reply by: Professor HovasapianWed Jan 18, 2017 7:39 PM

Post by Neil Kottaon June 28, 2016

In example three how is PO3(3-) oxidized? I thought it gained electrons there making it a reduction reaction not an oxidation. Thank you

1 answer

Last reply by: Professor HovasapianFri Apr 1, 2016 2:50 AM

Post by Sazzadur Khanon March 30, 2016

On example 3, which direction is the electron flow and why?

1 answer

Last reply by: Professor HovasapianTue Dec 2, 2014 2:42 AM

Post by Long Tranon November 30, 2014

Hi Professor,thank you for the lecture, i understand the process to break the reaction into 2 half rxn. However, i have difficulty to do the half rxn of this rxn. i hope you can helpO2(g) + 4H+(aq) + 4Fe2+(aq) -----> 4Fe3+(aq) + 2H2O(l)

2 answers

Last reply by: Christian FischerThu May 15, 2014 1:29 AM

Post by Christian Fischeron May 12, 2014

Hi raffi. Great lecture on redox chemistry - I really needed to refresh this. A quick question: When you add OH(-) at about 11.30min, do you allways add the same amount of OH(-) as the amount of H(+) added on each side in order to neutrulize the H+? And do you always add the same amount of OH(-) on the left and right side of the reaction?

Kind regards Christian

3 answers

Last reply by: Professor HovasapianFri Sep 27, 2013 5:28 PM

Post by Nithist kanjanavikaion September 22, 2013

Excuse me. In min 12:26, why does the left side of the equation has PO4^3-?

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Galvanic Cells

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Transcription: Galvanic Cells

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Last time, we introduced the notion of oxidation-reduction chemistry as a sort of an entry into our discussion of electrochemistry.0004

We talked about the balancing of an oxidation-reduction reaction (otherwise known as a redox reaction) under acidic conditions.0012

So, today I am going to give you an example of how to balance it under basic conditions, because as it turns out, reactions--for redox, they can either happen under basic, neutral, or acidic conditions.0020

Let's just go ahead and jump into the example, and we will go through what it is that we are doing as we go along.0033

OK, so the first example: we want to balance the following reaction under basic conditions.0040

And "basic conditions" just means hydroxide; so, somewhere along the way, we are going to add hydroxide to see...because remember, in "under acidic conditions," one of the things on one side or the other of the reaction was some hydrogen ion; that is the acid.0062

Under basic conditions, your base is going to be the hydroxide ion.0077

If you end up doing something under basic conditions, and you realize that there is no hydroxide anywhere, something is wrong; something happened.0082

OK, under basic conditions: So we have: PO33- (this is phosphite, by the way) + MnO4- goes to MnO2 (this is a solid) + PO43-.0091

These ions are all going to be aqueous; that is the whole idea behind ions--these are aqueous solutions.0115

Sometimes, though, you are going to end up with some solids or some gases, and we will specify whether they are solids or gases.0119

But, I'm not going to go ahead and put "aq" underneath each of these; when you are dealing with ions, you are talking about an aqueous solution (water); that is it.0126

OK, well, clearly, this is an oxidation-reduction reaction, because here, the phosphorus (PO3) goes to PO4 (right?--phosphorus becomes oxidized).0134

The oxidation states--you can confirm this by doing oxidation states: you know, let's just go ahead and do it while we are here.0145

Oxygen is a -2 oxidation state; there are 3 of them, for a total of -6; -6, plus what number, will give us a +3?0151

So here, phosphorus is a +3 oxidation state; over here, oxygen is -2; there is a total of 4 of them, so the total charge is -8; -8, plus what, will give us a -3?0163

Remember, the sum of all of the oxidation states of all of the individual atoms add up to the charge on the species.0175

In this case, the charge on the species is 3-, so this is 5+; so this is oxidized from a +3 to a +5--it loses two electrons.0183

Well, manganese--here, this is a +7, and this is going to end up being a +4; here, it gains three electrons.0193

So, the electrons don't balance; but we'll work that out as we go through.0203

Here are the steps for balancing an oxidation-reduction reaction under basic conditions.0206

OK, step 1 (you are going to like this): do the same thing as you did for acid conditions.0213

OK, step 2: once you do that (once you balance the reaction under acidic conditions), now, add equal amounts of OH- to each side of the reaction to balance out the H+--because, if you do it under acidic conditions--the first part--you are going to have some H+ left over.0238

Well, we want to neutralize that H+ with an equal amount of OH-, but we want to do it on both sides, so that one side ends up with water and the other side ends up with the hydroxide.0280

Step 3: eliminate H2O if it occurs on both sides (which we do anyway; any time we have two species on the same side of an equation, we cancel them out like we do in algebra, if it occurs on both sides).0291

We have a 3- and a 0, for a total of 3- on the left; we have 3- and 2+ on the right...so 3-, plus 2, is a -1; so we have -1 on the right, and we have -3 over here; so here, we need to add two electrons.0481

OK, now the next thing (again, we are still in step 1) we do is: we want to be able to cancel the electrons, so we have to multiply each equation, or one equation, by integers, to make sure that the electron count equalizes.0512

3 electrons on the left; 2 electrons on the right; so I'm going to multiply this equation by 2 (let me do this in red)--everything by 2--and this equation by 3.0526

OK, and now I'm going to rewrite these two equations: so I'm going to get 2 MnO4- + 8 H+ + 6 electrons goes to 2 MnO2 + 2 H2O...I'm sorry, 4 H2O (because it's 2 times 2, right?--yes, even we make mistakes).0540

And this one down here becomes: 3 H2O + 3 PO43- goes to 3 PO43- (my apologies; this is PO33-; wow, we have superscripts; we have subscripts; we have coefficients--all kinds of symbols going on here...where are we?) + 6 H+, and 6 electrons.0570

We have balanced each half-reaction; we have added; we have canceled the species; now, we are going to go to step 2--we are going to have to add hydroxide to both sides.0658

Let me rewrite the equation again: so, I have: 2 H+ + 3 PO43- + 2 MnO4- goes to 3 PO43- (the order doesn't matter, so don't feel like it has to be in a certain order; it doesn't) + 2MnO2 + H2O.0665

OK, so we have that; now, our second step was: we want to add an equal amount of hydroxide to cancel out the hydrogen ion.0694

We have two hydrogen ion over here, so I'm going to add 2 hydroxide ions...+ 2 hydroxide ions, to both sides.0701

OK, when we combine this, the H+ and the OH- combine to form 2 H2O + 3 PO43- (I'm going to write everything...you can do this without writing everything again and again, but I want you to see the steps) → 3 PO43- + 2 MnO2 + H2O + 2 OH-.0714

Well, now we have 2 H2O here; one H2O; so that H2O goes with that H2O, leaving only one water.0740

Our final is one water on the left, plus three phosphite ions, plus two permanganate ions, goes to three phosphate ions, plus two manganese dioxide (this is a solid), plus two hydroxides.0746

Here we go: this is our final balanced equation for basic conditions.0768

What that means is this--so let me tell you what this means, because there is a lot of symbolism here: This means that, if I put together--if I mix--some phosphite ion with some permanganate ion, the manganese is going to oxidize the phosphorus.0774

The MnO4 is going to turn into manganese dioxide; it's going to be a solid; it's going to drop to the bottom of the beaker if I just mix these two.0793

The phosphite is going to turn into phosphate; it's going to grab one of the oxygens that comes from the water, or maybe just one of these oxygens here.0800

There has to be some water floating around for this to happen; that is why this water is here.0809

OK, it's a basic oxidation-reduction reaction; this is how we balance it under basic conditions.0828

Good; OK, now there is one more that I would like to show you; this one is actually really, really easy.0835

I'll call this one Example 2 (let's go back to blue): Balance Zn2+...if I take zinc ion, and if I add some aluminum metal to that zinc ion solution, I end up with aluminum ion solution, and all of a sudden, zinc metal starts to show up in the solution.0841

So, in other words, aluminum dissolves, and zinc shows up; it will happen.0866

This is the oxidation half-reaction; these electrons are not equalized--we need to multiply them.0944

Now notice, there is no...1 Zn, 1 Zn; 1 Al, 1 Al; I'm still using the same process: balance the oxygens with H2O...there are no oxygens; balance the hydrogens...no hydrogens; balance the charge...we did--we added the electrons.0949

OK, everything is good; now, we just multiply by integers to equalize the electron count.0964

I'm going to multiply this equation by 3, and I'm going to multiply this equation by 2.0969

So, here is what happens: if I have some solution of, let's say, zinc sulfate--I have a bunch of zinc 2+ floating around and SO42- floating around--if I take a piece of aluminum and I actually drop it in here--just take a solid chunk of aluminum and I drop it in there--guess what is going to happen.1016

The aluminum is going to melt, and where the aluminum is melting, zinc is going to show up--because this says that it will.1036

We will get to (in the next lesson) how we know that this will happen; we are going to assign numbers to these things and decide, if I mix this with this, which one is going to melt and which one is not.1049

We will decide later, but right now, we are just working with balancing the reactions; but I want you to see what "balancing reaction" means.1061

Zinc metal--3 atoms of zinc metal--react with 2 (zinc ion; I'm sorry--this is 2+) atoms of aluminum metal; they rip electrons away from the aluminum--3 electrons from each aluminum atom; each zinc takes 2 of those electrons to become zinc metal, and it leaves the aluminum metal to go into solution.1069

Now, as we said, if I take some iron ion solution--iron 2+ ion solution--and I mix it with some permanganate solution, and if I add some acid to get things going, this reaction is going to take place.1153

The manganese is going to steal electrons from these iron atoms; it's going to become manganese 2+; it is going to release those oxygens.1170

Those oxygens are going to bind with the hydrogen in a shift of electrons being moved around among all these species; four water molecules are going to form, and the iron (2) is going to turn into iron (3).1178

If I take that and that--so I have this solution of MnO4- with a little bit of H+ thrown in for good measure, and here I have a solution of Fe2+...1203

It could be iron sulfate, iron chloride...we don't really care about the anion; it's the cation that is important in this particular case, and here, this could be potassium permanganate, calcium permanganate...it doesn't really matter--here it's the anion that matters, so we're interested in the species that are involved in the chemistry.1219

If I put these two solutions together, and if all of a sudden, I decide to take some metal wire (like some platinum wire or something), and if I connect them with a wire, what is going to happen?1236

Well, as it turns out, this reaction will take place; and the reason it will take place is because now, the electrons...before, when they were in the same beaker, the electrons just transferred face-to-face; here, if I connect them with a wire, the electrons have a path.1247

In other words, the manganese still wants the electrons that iron has to give up; therefore, it will take them from the iron; but the electrons will actually travel through the wire.1265

This oxidation-reduction reaction will take place, and it will take place spontaneously, because as written, this actually is a spontaneous reaction (and we will discover why later).1277

But, that is the whole idea; so, the oxidation reaction still takes place, but now, the electrons are passing through the wire.1287

Well, whenever we have electrons passing through a wire--whenever we have a current--we can use that current to do work.1298

We take an oxidation-reduction process as written; we separate the oxidizing agent and the reducing agent (we don't even need to call them that--we can just say we separate one species from the next); and we connect those two with a wire.1308

When we do that, we actually create a current; electrons spontaneously start to flow from here to here.1324

Iron gives up its electron; the electron travels through the wire; it hits the manganate ion; the manganese takes the electrons, one at a time; and once it has taken a certain number of them (5 of them, in this case), it will turn into manganese 2+.1332

So, in this compartment, manganese 2+ will start to show up; in this compartment, as iron 2+ gives up its electron, it turns into iron 3+.1347

A salt bridge is just a piece of glass tubing where, inside, there is a strong electrolyte (it could be sodium chloride, potassium chloride, calcium chloride); and what it does is: as electrons flow, and negative charge...well, let me draw it out, and you'll understand; otherwise, it's not going to make sense.1486

We have a little wire running like that; I haven't drawn in any things called electrodes, but I'll draw them in just a moment.1512

We have MnO4-, and we have H+; we have Fe2+, and we have this thing (actually, I'm going to write my MnO4- and H+ over here)--I have this tube, this glass tube, shaped like a U.1521

Inside, there is a strong electrolyte--in other words, a salt in a Jell-O-like matrix, in a soft matrix; and here is what happens.1544

When electrons flow this way, negative charge gathers over here; positive charge gathers over here; but as negative charge gathers over here, the positive ions of the strong electrolyte flow into the solution to balance the negative charge.1556

And here, as positive charge gathers on this side, the negative ions flow into solution here from the electrolyte, to balance the charge.1572

What this salt bridge does: it allows the passage of free ions to make sure that the charges are always balance, so there is not a buildup of charge; that takes a lot of energy to keep the charge separation.1581

Now, I'm going to draw the version that adds the porous disk in it.1623

In this particular case, you actually separate the solutions--they are almost in contact with each other, but there is this disk in between.1627

The only difference is that it allows the passage of these ions that balance charge.1634

For all practical purposes, this salt bridge and this porous disk shouldn't really concern us; the only thing you should know about them is that they are there simply to allow the passage of ions to balance charge as electrons flow.1640

MnO4-, H+, Fe2+...there are other ions floating around, positive and negative, that allow the flow of ions across this way and across that way, to make sure that, when electrons are flowing like that, negative charge builds up here; positive charges will flow across the ion, the porous disk, to balance.1659

And here, as positive charge builds up, negative ions will pass across the disk; water will not; the ions will, to balance the charge; that is all that is going on.1680

OK, so let's do this one as an example--I think that will be the best.1698

So, we'll do Example 3: OK, in the reaction of phosphite, of PO33-, with MnO4-...which...I'll remind you what that was: we have H2O + PO33- + MnO4- goes to 3 PO44- + MnO2, solid, plus 2 hydroxide; remember, this was in basic conditions...1705

So, for this example, we would like you to draw the galvanic cell, and identify the components.1749

OK, so now, we are going to get a little bit of vocabulary work on what a galvanic cell is and what each of these things is called.1766

OK, so I'm going to draw it with a porous disk; I think it's just easier to draw than with a salt bridge.1773

That is pretty much the standard nowadays--to use porous disks; salt bridges are sort of a little further back in time.1778

OK, so we have: there we go--phosphite is the thing that is oxidized; permanganate is the thing that is reduced (it is actually the manganese that is reduced, and it's the phosphorus that is oxidized--oxygen is still -2 oxidation state).1785

We are going to have a wire, and we are going to have, at the ends of the wires...we have these things called electrodes; and electrodes are usually broad pieces of metal (they could be long; sometimes they are not long, but they are generally flat); they are used to increase the surface area.1814

So, because you have this broad piece of metal, it is more in contact than actually just a thin wire is; so there is more opportunity for reactions to take place, like that.1832

If you had just a wire, it's only this much; but if you had this, it's all the space available on the...so these things are called...electrodes.1842

OK, they are the things at the ends of the wire that are actually dipped into these liquid solutions; they are called electrodes.1851

OK, and in this particular case, they are actually platinum electrodes; more often than not, platinum is used.1855

Now, the reason...we will actually talk a little bit about why we use these platinum electrodes in some of the other galvanic cells that we draw in just a little bit.1861

We are actually going to use electrodes made of the metals themselves, that are being oxidized and reduced.1883

But, in this case, because none of the species that is actually being oxidized or reduced is a metal in and of itself, it can't be used as an electrode.1888

We have to use some inert material--so basically a scaffold--someplace where the reaction can take place.1897

You notice: this is an aqueous ion; this is an aqueous ion; in order for them to react, they have to somehow come together.1904

Well, the surface of a nice, inert metal like platinum allows them a place for them to conduct business, if you will.1911

That is really all these electrodes do: it allows these aqueous ions to conduct their business, and it gives them a place to do that.1918

OK, so let's see what happens: that is the platinum electrode; let's go ahead and put...for the time being, it doesn't really matter where we put our solutions.1926

PO33-...I'll put my phosphite solution there; and I'll put my permanganate solution over here (MnO4-), and we also have some hydroxide, right?--that is over here.1937

Here is what is going to happen: we know that phosphite is oxidized; we know that permanganate is reduced; well, that means the thing that is oxidized...electrons are going to flow...so I'm going to write...1955

Here is what happens: an electron is stolen from phosphorus, travels through the electrode, and travels through the wire in this direction.1982

It comes down here, and at the interface, it jumps onto the manganese.1995

When enough of these electrons jump onto the manganese, what happens is that manganese dioxide starts to form on the electrode's surface, because this reaction is taking place on the electrode's surface.2000

This solution is in contact with this metal; where the ion meets the metal, that is where it releases its electron and lets it go.2016

As it releases its electron, there is water here; so it's going to turn into PO4; as this one is reduced, manganese dioxide is a solid, and it's going to start forming on the electrode, like that.2025

This is called the anode; anode is where oxidation takes place; this is what is important: where oxidation takes place, it's called the anode.2044

The best way to remember this: oxidation is a vowel; anode begins with a vowel.2059

That is the best way to remember that; that is still how I remember it.2064

Over here, this is called the cathode; the cathode is where reduction takes place.2067

In other words, it is where the electrons come to--where reduction takes place.2075

That is all that is happening here: something is being oxidized--it is giving up electrons; electrons are moving away from that area--they are being sucked over into here, because now, this one is going to take the electrons and use them.2081

This is being oxidized; this is being reduced; where oxidation takes place, this is called the anode; where reduction takes place, this is called the cathode.2093

So, notice what is happening here: zinc ion is turning into zinc metal; aluminum metal is turning into aluminum ion.2141

Because the metals are actually there as species, one of them is becoming a metal; one of the metals is turning into...you can actually use these metals as the electrodes.2149

You don't have to use platinum; you don't have to use some inert...the only time that you need some inert platinum electrode is when the things being oxidized and reduced are aqueous ions, and they stay aqueous ions.2162

This is the water level, by the way, so I should make it like this; this is just...that's right.2182

What this looks like is: there we go--for the time being, it doesn't matter where we put what, so how about if I put the aluminum...how about if I make this the aluminum metal, and if I make this the zinc metal.2187

That means here I have zinc ion in solution, and here I have aluminum ion in solution; and I actually may not have any aluminum ion in there yet.2205

What ends up happening is the following: zinc is being reduced; aluminum is being oxidized; well, here is the aluminum--it is oxidized; oxidation--anode, so this is the anode; this is the cathode.2213

Oxidation is where electrons are lost; that means this aluminum metal--electrons are being pulled from the aluminum metal, and they are traveling through this wire.2229

When they arrive here, they attach themselves to zinc ion that is touching the metal surface, and zinc metal starts to form on top of the zinc electrode.2240

That is why, because it's a metal, we can actually use it as the electrode; so zinc metal starts to form, and this aluminum starts to break down; the aluminum starts to dissolve.2257

When aluminum gives up its electrons, it turns into aluminum ion; so now, a whole bunch of aluminum ion is going into solution.2272

The metal is turning into the ion; here, the zinc ion is turning into the metal.2281

OK, now, let's see: let me write down: When the species being oxidized is a metal and/or the species being reduced becomes a metal (like this--aluminum is oxidized; it's a metal; zinc is the species...it's becoming a metal: we can go ahead and use those metals as the electrodes; you don't have to use platinum), then the electrodes can be the metals themselves.2288

OK, remember what we just said: aluminum metal--zinc wants the electrons; electrons disappear...an aluminum atom loses its electrons; the electrons flow this way.2362

When it loses its electrons, it becomes aluminum ion; therefore, aluminum ion goes into solution.2378

Zinc is forming: at this interface where the zinc metal is in contact with the zinc ion solution, the ions grab the electrons.2388

The minute they grab the electrons, the zinc 2+ becomes...they grab 2 electrons; they become zinc 0.2397

It becomes zinc metal; so zinc metal starts to form on top of the zinc metal electrode.2403

This starts to dissolve; now, this process right here--this metal dissolving--this is what happens when a metal dissolves in acid.2408

This is an oxidation-reduction reaction; as it turns out, when you drop a metal into acid (like, for example, if you drop zinc into acid or aluminum into acid), what is going to happen is an oxidation-reduction reaction.2419

So, I want you to be able to see this: so this--when metals dissolve in acid (not all metals do), this is an oxidation-reduction reaction, not an acid-base reaction like you might think.2434

Just because an acid is involved, in this particular case it doesn't mean it's an acid-base reaction.2472

Well, here is what is happening: if I take zinc metal, and I drop it into acid, zinc gets oxidized.2476

In other words, zinc loses 2 of its electrons, and it turns into zinc ion, plus two electrons.2482

That metal--that zinc metal--dissolves; and it is dissolving because the metal is turning into zinc ion, and the zinc ion is going into solution.2494